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Let’s consider the matrix

1.jpg

It has 2 rows & 2 columns

So, we write the order as

Order of a Matrix - Part 2

And,

Β  3, 2, 1, 4 are elements of matrix A

Β 

We write the matrix A as

Order of a Matrix - Part 3

Where

a 11 β†’ element in 1st row, 1st column

a 12 β†’ element in 1st row, 2nd column

a 21 β†’ element in 2nd row, 1st column

a 22 β†’ element in 2nd row, 2nd column

Β 

So,

Β  Β  a 11 = 3

Β  Β  a 12 = 2

Β  Β  a 21 = 1

Β  Β  a 22 = 4

Β 

For matrix

Order of a Matrix - Part 4

Β 

It has 3 rows & 2 columns

So, the order is 3 Γ— 2.

Β 

We write matrix B as

Order of a Matrix - Part 5

Β 

Similarly,

Order of a Matrix - Part 6

Β 

Create a 4 Γ— 3 matrix where elements are given by

a ij = i + j

Β 

A 4 Γ— 3 matrix looks like

Order of a Matrix - Part 7

Now,

a 11 = 1 + 1 = 2

a 12 = 1 + 2 = 3

a 13 = 1 + 3 = 4

a 21 = 2 + 1 = 3

a 22 = 2 + 2 = 4

a 23 = 2 + 3 = 5

a 32 = 3 + 2 = 5

a 33 = 3 + 3 = 6

a 41 = 4 + 1 = 5

a 42 = 4 + 2 = 5

a 43 = 4 + 3 = 7

Β 

So, our matrix is

Order of a Matrix - Part 8

Β 


Transcript

A = [β– 8(3&2@1&4)] 2 Γ— 2 Rows Column And, 3, 2, 1, 4 are elements of matrix A A = [β– 8(π‘Ž_11&π‘Ž_12@π‘Ž_21&π‘Ž_22 )] B = [β– 8(3&2@1&4@5&3)] B = [β– 8(3&2@1&4@5&3)]_(3 Γ— 2) Matrix Order [β– 8(9&5&2@1&8&5@3&1&6)] 3 Γ— 3 [β– 8(1&2&5&8&π‘₯&𝑧)] 1 Γ— 6 [β– 8(5@9@3@𝑦@tan^(βˆ’1)⁑π‘₯ )] 5 Γ— 1 [β– 8(sin⁑π‘₯&cos⁑π‘₯&tan⁑π‘₯&cot⁑π‘₯@sin⁑𝑦&cos⁑𝑦&tan⁑𝑦&cot⁑𝑦@sin⁑𝑧&cos⁑𝑧&tan⁑𝑧&cot⁑𝑧 )] 3 Γ— 4 A = [β– 8(π‘Ž_11&π‘Ž_12&π‘Ž_13@π‘Ž_21&π‘Ž_22&π‘Ž_23@π‘Ž_31&π‘Ž_32&π‘Ž_33@π‘Ž_41&π‘Ž_42&π‘Ž_43 )] A = [β– 8(2&3&4@3&4&5@4&5&6@5&6&7)] A = [β– 8(2&3&4@3&4&5@4&5&6@5&6&7)]

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.